 In the last lecture we have been looking at the equation for the salient pole alternator consisting of one damper circuit in each axis and with that we have these equations are shown for these equations which are five equations, five voltages and five currents written in this manner where kd represents the d axis damper and kq represents the q axis damper. These equations we have then reduced it to an equivalent electrical circuit one for the d axis as a shown here and one for the q axis as a shown here you have the d axis equivalent the d axis is in fact excited by two sources one is your vd of s another is the field voltage which is vf the q axis on the other hand is excited only by one source that is vq from the stator and this branch represents the d axis damper winding this is the magnetizing branch the leakage inductance of the stator and the field leakage inductance similarly on the q axis you have the q axis damper and then the leakage inductance of the stator and the magnetizing inductance along the q axis these equations can then be deduced from the original equations that we wrote over there. Now we may define flux linkage along the direct axis in this manner and the quadrature axis in this manner and if we see the equations with this definition of flux linkage the direct axis stator voltage equation can be written in this form r x i x p x psi d of s – ? psi q of s and the q axis equation for the stator can be written as r s i q plus d by dt of psi q plus ? times psi d these two equations are a direct result of our definition of the d axis and q axis flux linkage with this definition however if we now look at the equivalent circuit that we have written here and you compare it with the d axis voltage equation that you have here then v d of s is given by r s x id of s and then you have the speed emf which is this term therefore whatever voltage appears here must then be p times id of s by similar arguments whatever voltage appears here is then p times psi q of s. So that is to say that the circuit enclosed in the green rectangle here with vf being supplied at this point delivers at this terminal p times psi d of s and this circuit delivers p times psi q of s. Now it is of use to us to determine expressions for p times id of s and p times psi q of s as a function of the excitations for the circuit part shown in the green rectangle for this d axis equal in circuit the excitations are then as far as the green the circuit portion enclosed in the green rectangle is concerned what you have is vf of s here and another input that is going here is id of s in this case the only input is iq of s. So in order to look at that circuit in greater detail that portion of the circuit in greater detail we then drew that portion of the d axis equal in circuit first which is then excited by your id of s and v of s and this is your output variable that we are interested in. So we wrote the output variable of interest that is in the Laplace domain it becomes s times id of s is then obtained by two inputs and therefore the relation between the output variable and the two input variables can be obtained by superposition. So for the first case we consider vf to be 0 that means it is a short circuit and therefore you relate p times id to id of s and for the second portion the input is vf and therefore id is made 0 that means open circuit so vf into another transfer function b represents this and for these two we then obtain these expressions related that is a of s and b of s and we started looking at this function a of s as we have written here. So simplified it gives the numerator and denominator and we took up the denominator alone first for detail study and by looking at that expression of the denominator we found that it could be decomposed into a term rkd into rf multiplied by a second order term which is in this form and from that we identified the various t1, t2 and t3 these three terms as follows. Now we start we need to start looking at the numerator function so a of s is again given by the same expression but now we start looking at the numerator term. The numerator term can be obtained as this denominator multiplying s times the leakage inductance of the stator and that is then given by s times ll of s multiplying rkd into rf is what is coming here plus s times llkd into rf that term is here and then llf into rkd that term is here so this is an expansion of the first two terms here and then you have an s square term lmd okay lkd llf lkd into lf and l so that comes from here llkd into lf please make that change here so this is really llkd and then what we have is the s square term from here that is lmd into llf comes from here and llkd comes from here this term is llkd so let us make that change so this is llkd and then you have s times lmd that is the first order term so lmd into rf and lmd into rkd so this first row then accounts for the product of lls and the denominator here and then we need to account for the numerator term that comes out as s times lmd that is here and then the product of these two that is rkd into rf and then s term which is llkd into rf and llf into rkd and then the second order term llf into llkd so that is what comes here now this expansion can further be grouped we will group all the s exponential terms together so s cube, s square and s terms s cube term comes from the product of this s square and this s similarly this s square and this s and similarly these two terms also so if we take the first s cube term you have lls multiplied by llkd into llf so that is also llf this is llf and then so that is the first term and then you have llslmd multiplied by llf plus llkd so lls is what you have here that gets multiplied by this lmd that is this term and then you have llf that is coming here and then llkd comes from here so that is the second term and then the third you have s square that is s cube this term lmd into llf into llkd so that forms the three terms comprising of the s cube there and then we go to the s square terms s square terms come from the product of s terms here and this one also and then this s square this will also give another s square so s square gives lls multiplied by llkd rf that is there here llf into rkd that comes here and then you have this one lmd multiplying lls that is here multiplied by your rf plus rkd so that comes over here and then the third s square terms comes from this part lmd multiplying llf rkd and llkd and rf and then finally you have the s term that would come from multiplying this s by the terms s power 0 here that is rkd into rf so s times lls rkd rf comes here and then lmd rf rkd that comes here so these are the three terms and now we try to simplify them further we take the s term first s term is there you see that rf rkd is common in both these cases so we take it out so s rkd rf multiplied by then lls and lmd lls plus lmd so this is the first term and then if you look at the s square terms now lkd you have lls lkd into rf and here you have lmd lkd into rf so we take this lkd into rf common and what you are left with is lls plus lmd that comes here and then you have llf into rkd here and here also you have llf into rkd so that if you take out your left with lls plus lmd again so that comes here and then the third term is what is remaining that is lls into lmd into rf plus rkd which is nothing but this term which comes here as it is so what we have done is regroup them such that you have this lls plus lmd term identifying terms which are the same as lls plus lmd in all these different s terms and then when we go to sq you have lls lkd lf and lmd lflkd so that again you have lls and llkd if we take it out you are left with lls plus lmd and similarly then the next term which is lls lmd into llf plus llkd so in this manner we segregate the terms and group them together and then what we find is that your lls plus lmd occurs in many places so it may be worth taking it out separately and we take rs rf rkd also outside and since all these terms have at least one s term so we take that also out and what we are left with so lmd plus lls and rkd rf has been taken out so the first term now becomes one within the bracket and then let us look at the second term from this sq we have taken one s outside and lmd plus lls has also been taken outside so what we are left with is s times llkd into rf divided by rkd into rf that is what would result for the first term and then similarly s times when you go to the second term we have taken lls plus lmd out and therefore you have llf rkd by rkd rf and then you have lls lmd by lls plus lmd multiplied by rf plus rkd divided by rf rkd multiplied by s of course that is what you have for the s term may be what we could do is to group all these terms together since all our s terms we can put that around the common bracket so that is all s term and then the sq term what we had earlier since we are taking one s outside so you are left with sq what you have is llkd llf by rkd rf and then you have lls lmd by lls plus lmd multiplied by llf plus llkd by rkd rf so that is your sq term and in that manner so if we take the srkd rf outside along with lls plus lmd what remains within the bracket is this group here again what we can see is in these terms rf is common in numerator and denominator that can be cancelled here rkd is common that can be cancelled so each of these terms now is an l by r term and that can therefore be identified in some manner that we will see so with this then what we have written is in this form so llkd by rkd that is the first term then llf by rf so that is what we have here llf by rf and then you have product of two inductances divided by their sum lls lmd by lls plus lmd multiplied by this and then as the sq term you have llkd lf by rkd rf so that is what we have here and here you have lls the same term coming here and llkd by rkd rf so that is again what we have here so in this manner one can simplify now this term can further be written as lls lmd by lls plus lmd multiplied by rf by rf rkd we are splitting this into two parts plus lls lmd by lls plus lmd multiplied by rkd by rf rkd and similarly what is written here can be split as lls lmd by lls plus lmd multiplied by llf by rkd rf plus lls lmd multiplied by llkd divided by lls plus lmd multiplied by rkd rf having split it in this form what we can now see is this term can be reduced further in this manner so slmd plus lls is here rf rkd which we had written there earlier now has been written here one is there as it is and then s times now you see that in this term that is we have expanded this into two terms in this you have rf here and rf here that is removed so there are now two terms with the denominator as rkd you have llkd by rkd and lls lmd by lls plus lmd multiplied by rkd so that we can take out one by rkd from that and write it as llkd plus lls lmd by lls plus lmd this is one term and then you have a second term where the denominator is rf so here also then we have another term with the denominator as rf and therefore one over rf divided by ll multiplied by lls plus lls lmd by lls plus lmd so the s term can further be split into two terms in this manner the second s square term what we have done is you have llf llkd by rf rkd and then so you get one llkd here and then you have this term lls lmd by lls plus lmd so this also has an llf by rkd rf therefore if we combine this and this term we can take out llf by rf rkd we are left with llkd plus this term coming over here and then the term that is there here is lls lmd llkd divided by lls plus lmd multiplied by rkd rf this term then comes over here. Now the denominator can further be expanded in this manner if you take rf rkd also outside here s square by rf rkd then you have llf llkd that comes over here and then you have lls lls lmd by lls plus lmd and then you have this term here which further can be simplified as now we can take this term and this term together take out llkd common from both and then you are left with llf that comes here and this remaining term comes here the third term is then here therefore with this simplification what you get is slmd l plus lls and then note that I have left out rf rkd in the numerator for a particular reason we will come to that and then 1 plus s times these two terms so that comes out as it is now in the s square term what we do is you have already s square rf rkd s square by rf rkd here now we take out this term as well llf plus lls lmd this also if you take it out of this bracket then you are left with llkd this is actually llkd plus this term divided by your llf plus lls lmd and so on so when that is simplified you get this term here that is let us do that small work out so what you have here the last term is llf lls lmd divided by lls plus lmd since we are taking out llf plus lls lmd by lls plus lmd we have to multiply it by lls plus lmd divided by the numerator of this term which is nothing but llf lls plus llf lmd plus lls lmd and this cancels with this therefore what you are left with is lls lmd llf that is what is here divided by the denominator which consists of products of two inductances so that is there here. So having written it in this form we find that the numerator can be written in a more condensed form as s times lmd plus lls is here and then you have 1 plus s times each of these is an l by r ratio so you have a time constant here you have a time constant here we call that as t5 plus t4 and then you have one time constant here multiplied by another time constant we call that as t4 t6 note that this l by r ratio l divided by rf is the same as what is occurring here so we call that as t4 the other time constant is t6. Now let us come to the item that I said I left out that is we actually have an rf rkd here and therefore this should have been multiplied by an rf rkd we remember if you look back at the numerator expression here the numerator also has an rf rkd and now in the denominator also we have an rf rkd therefore this gets removed with the rf rkd in the denominator and therefore what we are left with is this term where then we can write t5 as 1 by rkd into llkd plus lls lmd divided by lls plus lmd this is t5 and then you have t4 as 1 over rf multiplied by llf plus lls lmd divided by lls plus lmd and then t6 is nothing but 1 by rkd multiplied by llkd plus lls lmd llf divided by lls llf plus lls lmd plus lls lmd. So the three numbers that we land up with are these t5, t4 or three expressions are these. So now we have derived an expression for the factor a of s which is the first term in the expression here. So we wrote ?d of s, s times ?d of s as a id of s plus b times v of s what we have now done is simplified the expression a of s into definite identifiable simple factors and now we need to look at this b of s. So b of s we have seen could be written like this it is the same expression what we have written down there. So the denominator again can be written as rf plus slf plus slmd into rkd plus sllkd this these three terms can be grouped as rkd plus sllkd plus s times lmd. So let us try to simplify this this can then be written as s times rkd plus sllkd into lmd note that this denominator is the same as this denominator and therefore if we now multiply the denominator out what you have is rf plus slf this should actually be sllf that is the leakage inductance of the field this multiplied by rkd plus s times llkd plus rf plus sllf multiplying s times lmd plus s times lmd multiplied by rkd plus s times llkd this denominator later on is cancelled with that denominator. So this is the expression that we have this denominator we have already seen this where have we seen this if you look at the expression for a of s if you look at the expression for a of s and see the denominator of a of s the denominator of a of s is this term and this is exactly what we have rkd plus slllkd into rf plus slf that is the term here rf plus slf rkd in the so on and then you have rf plus sllf into slmd so that is exactly what you have here and the third term is again what is available there and therefore we already know what this term simplifies to this term simply gives us the denominator of s in this expression simply gives us rf rkd multiplied by 1 plus s into t1 plus t2 plus s square multiplied by this term that is t1 t3 this is as far as the denominator is concerned. So let us look at the numerator now the numerator of s is nothing but s that is s times lmd into rkd plus sllkd and we take rkd outside and therefore this is s times lmd into rkd multiplied by 1 plus s times llkd by rkd and therefore this function b of s can now be written as on the numerator you have a term 1 plus s tkd where if we define tkd as llkd by rkd so s times tkd and then you have slmd rkd slmd is there on the denominator you have rf rkd and this rkd cancels with that rkd and therefore what we are left with is slmd by rf and therefore the expression b of s now looks like this first order numerator by the same second order denominator first order numerator plus into another s term. So this is as far as the d axis is concerned now we need to look at the q axis in the q axis the equivalent circuit that we had was you had the leakage inductance of the stator and then the magnetizing inductance of the q axis lmq and then you have the q axis damper branch that is there here this is rkq llkq and what is going into this is your current iq of s and what we are interested in is this voltage which is s times psiq of s and therefore we can write this voltage s times psiq of s as this current multiplied by the overall impedance that is offered by this circuit the impedance that is offered by that circuit is nothing but the impedance of this which is s times lls plus the impedance offered by the combination of this branch and this branch and that is nothing but product of these two impedances multiplied by the sum of these two impedances and therefore you get this term here again we can look at the numerator and denominator separately the numerator is going to consist of an s square term which comes from the product of these two and these two terms so you have s square lls multiplying lmq plus llkq that is there here and then you have s square lmq llkq that is coming here and then you have the s term that comes basically from these two and s term multiplying this constant so you have s plus lls rkq that is there here and then lmq rkq that is there here and what we could do with this is you can see that rkq is common so we take that out then we are left with lls plus lmq so s into lls plus lmq into rkq if we take as common out of the entire numerator you have s into lmq plus lls multiplied by this rkq has been taken common outside so that is 1 plus s multiplied by now here if you take llkq common you are left again with lls plus lmq so that is what we have taken out here and therefore what will come within this bracket is only llkq and then you are left with this other s term which is lls lmq since we are taking lmq plus lls outside this gets divided by lmq plus lls and since we are taking rkq common you have to now divide by rkq here as well and therefore the numerator is simplified into this form as far as the denominator is concerned it is fairly straight forward denominator can be written as rkq into 1 plus s times lmq plus llkq divided by rkq this is your denominator which comes straight away from the expression here so now we have an expression for numerator and denominator both have an rkq that is taken common outside so this rkq cancels with that rkq and therefore what we are left with is s times lq of s note that the magnetizing inductance plus the leakage inductance is nothing but the quadrature axis self inductance of the stator so that is what is there here 1 plus st7 you here you have another l by r term so t7 divided by 1 plus st8 and therefore what we have is t7 equals 1 by rkq into llkq plus lls lmq divided by lmq plus lls and then you have t8 as lmq plus llkq by rkq so this term is nothing but the quadrature axis damper time constant that you have now let us take a look at all the t1 to t8 that we have derived before we proceed ahead let us go back so t1 is as we have said is nothing but the field time constant this is the field inductance divided by the field resistance so that is again straight forward t2 is the d axis damper time constant which is nothing but the inductance of the damper winding divided by the resistance of the damper winding so that is also straight forward if you look at t3 what happens to how does t3 come about let us take a look at our equivalent circuit which I will redraw here you have the leakage inductance of the stator and then the magnetizing inductance lmd and then you have the damper branch rkd and llkd and then your field part that comes here note that the time constant that we are seeing here as t3 can be obtained as follows if you assume that this side is open this is open and you short this side okay neglect the field resistance assume that we do not consider the field resistance now this term can be interpreted as follows what you get here is suppose you look at the resistance rkd and across the terminals of the resistance rkd if you want to find out what is the net inductance seen by this resistance what would you do you would have now this inductance connected like this therefore the inductance seen by this resistance is nothing but the inductance as appears as llkd and then you have two inductances lmd and this is llf this inductance is connected across the terminals of rkd and the value of this inductance is nothing but the parallel combination of your lmd and llf which is nothing but lmd llf by lmd plus llf this whole thing occurring in series with lkd so this is the total inductance that is seen and l by r is divided by rkd so it is important to note that this can be derived by considering the stator side to be open and looking at the inductance of the damper resistance I mean looking at the equivalent inductance seen by the damper resistance. Now let us look at the other time constants that we have so here you have some more time constants so let us look at these are still for the d axis so let us redraw that circuit here again then you have the damper and then you have the field so this is lmd this is lls rkd llkd llf and r so now let us look at your the inductance the number t5 the expression for t5 so that is llkd llkd plus lls lmd by lls plus lmd how does this come about suppose you assume that this side has been shorted and then you look at the equivalent inductance as seen across the terminals of rkd then what you would see is if this is open then lls lmd come in across each other so you get this combination of lls lmd by this plus this llkd so t5 can be obtained by considering the armature side to be short circuited and then looking at the equivalent inductance seen by the damper resistance t4 is then 1 over rf llf plus lls lmd by this this can be obtained by assuming that the damper does not exist if the damper is not there and you consider this to be shorted then the equivalent inductance seen across the terminals of rf will then be llf occurring in series with the combination of lls and lmd coming across each other and therefore this is the combination inductance plus lls so both t4 and t5 are then can then be visualized by assuming that the stator side is short circuited and then if you look at t6 what you have is llkd then lls lmd llf so now if you assume that this is shorted and this is also shorted and you look at neglect rf and you look at the resistance rkd and see the inductance that is seen by the resistance rkd in that case then that will consist of this llkd coming in series with a combination of three inductances coming across each other so you have llkd coming in series with one you have lls this comes this is shorted now lls and then you have lmd and then you have llf so the total inductance across these two terminals is nothing but llkd plus all this term this is one can derive this very easily so I will leave this for you to derive it from this combination so this is the total inductance as seen by the terminals at rkd and therefore this l by r ratio is the time constant what is interesting to note is that all the three time constants are derived by or may be derived from circuits where you assume that the stator terminals have been short circuited for this case this is just lkd by rkd which is the damper part and the denominator is the same as what we have seen note again that the denominator terms when we explained how they can be arrived at we said that the denominator terms can be thought of as being open circuited and the numerator then comes out as being where the armature terminals can be considered to be short circuited so that is an interesting observation that we make and then we went to the q axis equivalent circuit where we have derived simplified expression for that consisting of one term t7 and t8 t7 you can see is llkq so that can also be as arrived at by saying that you short these two terminals then across rkq the inductance that is seen is llkq plus the combination of these two coming across each other which is what is this term so here again the numerator term t7 can be arrived at by considering that the stator is short circuited in the denominator if you open this the armature side then as seen by the resistance of the damper the inductance is nothing but llkq plus lmq which is exactly what you have here so here again what we find is that the denominator term t8 can be arrived at by assuming that the stator terminal is open so now that we have simplified these expressions let us see how this can be used to further analyze performance of the synchronous machine in the next class we will stop with this for today.